Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y^2 + 8y - 20}{y^2 + 6y} \times \dfrac{y + 6}{y - 2} $
Solution: First factor the quadratic. $z = \dfrac{(y - 2)(y + 10)}{y^2 + 6y} \times \dfrac{y + 6}{y - 2} $ Then factor out any other terms. $z = \dfrac{(y - 2)(y + 10)}{y(y + 6)} \times \dfrac{y + 6}{y - 2} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (y - 2)(y + 10) \times (y + 6) } { y(y + 6) \times (y - 2) } $ $z = \dfrac{ (y - 2)(y + 10)(y + 6)}{ y(y + 6)(y - 2)} $ Notice that $(y + 6)$ and $(y - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(y - 2)}(y + 10)(y + 6)}{ y(y + 6)\cancel{(y - 2)}} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $z = \dfrac{ \cancel{(y - 2)}(y + 10)\cancel{(y + 6)}}{ y\cancel{(y + 6)}\cancel{(y - 2)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $z = \dfrac{y + 10}{y} ; \space y \neq 2 ; \space y \neq -6 $